Math Problem Statement

derivative of (sinx - cosx)/(sinx + cosx)

Solution

To find the derivative of the function sinxcosxsinx+cosx\frac{\sin x - \cos x}{\sin x + \cos x}, we'll use the quotient rule. The quotient rule states that if you have a function of the form u(x)v(x)\frac{u(x)}{v(x)}, its derivative is given by:

(uv)=uvuvv2\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}

Here, let:

  • u(x)=sinxcosxu(x) = \sin x - \cos x
  • v(x)=sinx+cosxv(x) = \sin x + \cos x

Step 1: Find the derivatives of u(x)u(x) and v(x)v(x)

  • u(x)=cosx+sinxu'(x) = \cos x + \sin x
  • v(x)=cosxsinxv'(x) = \cos x - \sin x

Step 2: Apply the quotient rule

Now, substitute u(x)u(x), v(x)v(x), u(x)u'(x), and v(x)v'(x) into the quotient rule formula:

ddx(sinxcosxsinx+cosx)=(cosx+sinx)(sinx+cosx)(sinxcosx)(cosxsinx)(sinx+cosx)2\frac{d}{dx}\left(\frac{\sin x - \cos x}{\sin x + \cos x}\right) = \frac{(\cos x + \sin x)(\sin x + \cos x) - (\sin x - \cos x)(\cos x - \sin x)}{(\sin x + \cos x)^2}

Step 3: Simplify the numerator

Let's expand and simplify the numerator:

  1. Expand (cosx+sinx)(sinx+cosx)(\cos x + \sin x)(\sin x + \cos x): (cosx+sinx)(sinx+cosx)=cos2x+cosxsinx+sinxcosx+sin2x=cos2x+sin2x+2cosxsinx(\cos x + \sin x)(\sin x + \cos x) = \cos^2 x + \cos x \sin x + \sin x \cos x + \sin^2 x = \cos^2 x + \sin^2 x + 2\cos x \sin x Since cos2x+sin2x=1\cos^2 x + \sin^2 x = 1, the expression simplifies to: 1+2cosxsinx1 + 2\cos x \sin x

  2. Expand (sinxcosx)(cosxsinx)(\sin x - \cos x)(\cos x - \sin x): (sinxcosx)(cosxsinx)=sin2xsinxcosxcosxsinx+cos2x=sin2x+cos2x2sinxcosx(\sin x - \cos x)(\cos x - \sin x) = \sin^2 x - \sin x \cos x - \cos x \sin x + \cos^2 x = \sin^2 x + \cos^2 x - 2\sin x \cos x Again, sin2x+cos2x=1\sin^2 x + \cos^2 x = 1, so this simplifies to: 12sinxcosx1 - 2\sin x \cos x

Thus, the numerator becomes: (1+2cosxsinx)(12sinxcosx)=4sinxcosx(1 + 2\cos x \sin x) - (1 - 2\sin x \cos x) = 4\sin x \cos x

Step 4: Simplify the expression

Finally, the derivative is: 4sinxcosx(sinx+cosx)2\frac{4\sin x \cos x}{(\sin x + \cos x)^2}

This can also be written using the identity sin2x=2sinxcosx\sin 2x = 2\sin x \cos x as: 2sin2x(sinx+cosx)2\frac{2\sin 2x}{(\sin x + \cos x)^2}

Final Answer

The derivative of sinxcosxsinx+cosx\frac{\sin x - \cos x}{\sin x + \cos x} is:

2sin2x(sinx+cosx)2\frac{2\sin 2x}{(\sin x + \cos x)^2}

Would you like any further details or explanations?


Here are 5 related questions:

  1. What is the derivative of sinx+cosxsinxcosx\frac{\sin x + \cos x}{\sin x - \cos x}?
  2. How would you find the second derivative of sinxcosxsinx+cosx\frac{\sin x - \cos x}{\sin x + \cos x}?
  3. Can the derivative 2sin2x(sinx+cosx)2\frac{2\sin 2x}{(\sin x + \cos x)^2} be simplified further?
  4. What is the integral of sinxcosxsinx+cosx\frac{\sin x - \cos x}{\sin x + \cos x}?
  5. How do you apply the quotient rule to functions involving trigonometric identities?

Tip: Remember that trigonometric identities can often simplify the differentiation or integration process, making complex expressions easier to handle.

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Math Problem Analysis

Mathematical Concepts

Differentiation
Trigonometric Functions
Quotient Rule

Formulas

Quotient rule for differentiation

Theorems

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Suitable Grade Level

Advanced High School / College